zoukankan      html  css  js  c++  java
  • Sicily 2002. Feeding Time (广搜) 解题报告

    Description

    It's Bessie's feeding time, and  Farmer John is trying to decide where to put her. FJ has a farm that  comprises W x H (1 <= W <= 750; 1 <= H <= 750) squares and  is partitioned into one or more separate pastures by rocks both large  and small. Every pasture contains some grass and some rocks.
    Bessie is a hungry little cow and just loves to eat, eat, eat her  grass. She can move from any square to any other square that is  horizontally, vertically, or diagonally adjacent. Bessie can't cross the  rocks because they hurt her feet, and, of course, she can't leave the  farm. Bessie wants to know the maximum number of squares of grass that  she can eat.
    FJ has a map of his farm, where a '.' represents a square of grass,  and a '*' represents a rock. Consider this 10x8 map and a detailed  breakdown of the extent of each of its three pastures:

          ...*....**  |  111*....**   ...*2222**    ...*....**
    
          ..**....**  |  11**....**   ..**2222**    ..**....**
    
          ...*....**  |  111*....**   ...*2222**    ...*....**
    
          ...**.*.**  |  111**.*.**   ...**2*2**    ...**.*.**
    
          ***.**.***  |  ***1**.***   ***.**2***    ***.**.***
    
          ...**.*.**  |  111**.*.**   ...**2*2**    ...**.*.**
    
          ...*.*****  |  111*.*****   ...*2*****    ...*.*****
    
          ...***..**  |  111***..**   ...***..**    ...***33**

    Pasture 1 has 21 squares; pasture 2 has 18 squares; pasture 3 has 2  squares. Thus Bessie should choose pasture 1 with 21 squares to maximize  the grass she can eat.
    Input

    * Line 1: Two space-separated integers: W and H
    * Lines 2..H+1: Line i+1 describes field row i with W characters (and no spaces), each either '.' or '*'

    Output

    * Line 1: A single integer that represents the maximum number of squares of grass that Bessie can eat.

    Sample Input
                      
    10 8
    ...*....**
    ..**....**
    ...*....**
    ...**.*.**
    ***.**.***
    ...**.*.**
    ...*.*****
    ...***..**
    Sample Output
    21
    解题报告:
      这是一道典型的搜索题,可以使用广搜实现。题目意思大致就是找出一片相邻的最大面积。使用广搜基本方法即可,无需任何处理技巧。
    代码如下:
    1 #include<iostream>
    2 #include<queue>
    3 using namespace std;
    4 const int MAX=760;
    5
    6 struct node
    7 {
    8 int x,y;
    9 char gr;
    10 };
    11
    12 int counts;
    13 node nod[MAX][MAX];
    14 int visit[MAX][MAX];
    15 int bfs(int a,int b)
    16 {
    17 queue<node>que;
    18 node nod2;
    19 int i,j;
    20 que.push(nod[a][b]);
    21 visit[a][b]=1;
    22 counts++;
    23 while(!que.empty())
    24 {
    25 nod2=que.front();
    26 que.pop();
    27 i=nod2.x;
    28 j=nod2.y;
    29 if(nod[i-1][j].gr=='.' && visit[i-1][j]==0)
    30 {
    31 que.push(nod[i-1][j]);
    32 visit[i-1][j]=1;
    33 counts++;
    34 }
    35 if(nod[i+1][j].gr=='.' && visit[i+1][j]==0)
    36 {
    37 que.push(nod[i+1][j]);
    38 visit[i+1][j]=1;
    39 counts++;
    40 }
    41 if(nod[i][j-1].gr=='.' && visit[i][j-1]==0)
    42 {
    43 que.push(nod[i][j-1]);
    44 visit[i][j-1]=1;
    45 counts++;
    46 }
    47 if(nod[i][j+1].gr=='.' && visit[i][j+1]==0)
    48 {
    49 que.push(nod[i][j+1]);
    50 visit[i][j+1]=1;
    51 counts++;
    52 }
    53 if(nod[i-1][j-1].gr=='.' && visit[i-1][j-1]==0)
    54 {
    55 que.push(nod[i-1][j-1]);
    56 visit[i-1][j-1]=1;
    57 counts++;
    58 }
    59 if(nod[i+1][j+1].gr=='.' && visit[i+1][j+1]==0)
    60 {
    61 que.push(nod[i+1][j+1]);
    62 visit[i+1][j+1]=1;
    63 counts++;
    64 }
    65 if(nod[i-1][j+1].gr=='.' && visit[i-1][j+1]==0)
    66 {
    67 que.push(nod[i-1][j+1]);
    68 visit[i-1][j+1]=1;
    69 counts++;
    70 }
    71 if(nod[i+1][j-1].gr=='.' && visit[i+1][j-1]==0)
    72 {
    73 que.push(nod[i+1][j-1]);
    74 visit[i+1][j-1]=1;
    75 counts++;
    76 }
    77 }
    78 return counts;
    79 }
    80
    81 int main()
    82 {
    83 int i,j;
    84 int w,h;
    85 int large=0;
    86 cin>>w>>h;
    87 for(i=1;i<=h;i++)
    88 for(j=1;j<=w;j++)
    89 {
    90 cin>>nod[i][j].gr;
    91 nod[i][j].x=i;
    92 nod[i][j].y=j;
    93 visit[i][j]=0;
    94 }
    95 for(i=1;i<=h;i++)
    96 for(j=1;j<=w;j++)
    97 {
    98 if(nod[i][j].gr=='.' && visit[i][j]==0)
    99 {
    100 counts=0;
    101 counts=bfs(i,j);
    102 if(counts>large)large=counts;
    103 }
    104 }
    105 cout<<large<<endl;
    106 return 0;
    107 }
  • 相关阅读:
    Eclipse出现Access restriction : The type BASE64Decoder is not accessible due to restriction问题
    Struts学习总结(1)
    TCPMon使用总结
    AJAX入门学习总结
    JSON for java入门总结
    JSTL使用总结(2) fmt标签库和fn标签库
    JSTL使用总结(1) 核心标签库和SQL标签库
    JavaScript学习总结
    Ant学习总结
    C# .NET中如何使用GetCursorPos函数
  • 原文地址:https://www.cnblogs.com/lvye/p/1995668.html
Copyright © 2011-2022 走看看